﻿ 关于中微子质量属性与弱力关系的研究 The Research on Relationship between Neureinos and Weak Force

Modern Physics
Vol.07 No.05(2017), Article ID:22142,16 pages
10.12677/MP.2017.75023

The Research on Relationship between Neureinos and Weak Force

Zhengdong Huang

Hubei Zhouheiya Corporate Development Co., Ltd., Wuhan Hubei

Received: Sep. 1st, 2017; accepted: Sep. 16th, 2017; published: Sep. 25th, 2017

ABSTRACT

The main idea of the article is to refute the neutrinos carrying mass. Firstly, the three neutrino production processes to capture are found between neutrinos and decay; the weak force is associated; and then the relationship between neutrinos and weak force is researched. Corresponding to the analysis of four independent physical phenomena, a more complete neutrino is obtained which has no mass.

Keywords:Neutrino Oscillation, Neutral Keon Quality Error, Electron Magnetic Moment Anomaly, Weak Force, Weak Interaction

1. 引言

2015年的物理学诺贝尔奖授予了Takaaki Kajita和Arthur B. McDonald，以表彰他们对中微子振荡现象的发现。中微子振荡现象的发现证实了中微子是拥有质量的，从而将它从光影阑珊的世界拉入了世俗这个充满了约束与限制的囚笼。但笔者对这个结论持怀疑态度，疑点有三。

1) 反电子中微子的产生过程如下图1

2) 反电子中微子的运动过程如下图2

3) 反电子中微子的捕获过程如下图3

Figure 1. Pion decay modes

Figure 2. The motion of anti-electron neutrino

Figure 3. The trapped of anti-electron neutrino

2. 弱力势解析

2.1. 力荷及力荷类型假设

${\left(\underset{0\le i\le n}{\overset{n\in \left[0,3\right]}{\sum }}{e}_{i}\right)}^{2}=0$ (1)

2.2. 弱力势假设

$\begin{array}{l}{\stackrel{^}{V}}_{W}\left(r\right)=-\frac{8\text{π}}{3}{\stackrel{^}{e}}_{1}\cdot {\stackrel{^}{e}}_{2}\delta \left(r\right)\\ {\stackrel{^}{e}}_{i}={a}_{i}{\alpha }_{x}+{b}_{i}{\alpha }_{y}+{c}_{i}{\alpha }_{z}\\ {a}_{i},{b}_{i},{c}_{i}=\text{constants}\end{array}$ (2)

${\left({\stackrel{^}{e}}_{i}\right)}^{2}=0$ (3)

${\left({a}_{i}{\alpha }_{x}+{b}_{i}{\alpha }_{y}+{c}_{i}{\alpha }_{z}\right)}^{2}=0$ (4)

${a}_{i}=1,{b}_{i}=±i$ (5)

$\left\{\begin{array}{l}{\stackrel{^}{e}}_{1}={\alpha }_{x}+i{\alpha }_{y}\\ {\stackrel{^}{e}}_{1}={\alpha }_{x}-i{\alpha }_{y}\end{array}$ (6)

$\phi \left(r\right)=\mathrm{exp}\left(-r/a\right)$ (7)

3. 中性K介子的质量差

$\frac{{m}_{{K}_{±}}}{{m}_{{K}_{0}}}=\frac{1+0.5\mathrm{ln}\left(1-4\left(1-1/3\sqrt{3}\right)\alpha \right)}{1-0.5\mathrm{ln}\left(1+\alpha \right)}=0.9916780905$ (8)

$\frac{{m}_{{K}_{±}\mathrm{exp}}}{{m}_{{K}_{0}\mathrm{exp}}}=\frac{493.677±0.016}{497.648±0.024}=0.992020±\cdots$ (9)

$\stackrel{¯}{s}\stackrel{¯}{d}\text{ }\stackrel{¯}{d}⇔{\mu }^{+}$ (10)

Figure 4. The internal plane structure of kaon

${\mu }^{+}\to {e}^{+}+{\upsilon }_{e}+{\upsilon }_{\mu }$ (11)

$\begin{array}{c}\Delta E=E\left(F=1\right)-E\left(F=2\right)\\ =\frac{2}{3}{g}_{p}{g}_{e}\frac{{m}_{e}}{{m}_{p}}{m}_{e}{\alpha }^{4}{c}^{2}=5.88×{10}^{-6}\text{\hspace{0.17em}}\text{eV}\end{array}$ (12)

$\begin{array}{c}\Delta {E}_{1}=E\left(F=1\right)-E\left(F=2\right)\\ =\frac{2}{3}{g}_{\mu }{g}_{e}\frac{{m}_{e}}{{m}_{\mu }}{m}_{e}{\alpha }^{4}{c}^{2}\end{array}$ (13)

$\mathrm{Re}=\frac{3}{8}$ (14)

$\begin{array}{c}\Delta {E}_{2}=\Delta {E}_{1}×\frac{1}{2}\mathrm{Re}\\ =\frac{\mathrm{Re}}{3}{g}_{\mu }{g}_{e}\frac{{m}_{e}}{{m}_{\mu }}{m}_{e}{\alpha }^{4}{c}^{2}\\ =0.530054\left({10}^{10}\hslash {s}^{-}\right)\end{array}$ (15-1)

${m}_{{K}_{L}}-{m}_{{K}_{S}}=\left(0.5293±0.0009\right)\left({10}^{10}\hslash {s}^{-}\right)$ (15-2)

Figure 5. Hyperfine structure splitting

$\begin{array}{c}\Delta {E}_{A}=\Delta {E}_{1}×\frac{1}{2}\mathrm{Re}\\ =\frac{\mathrm{Re}}{3}{g}_{\mu }{g}_{e}\frac{{m}_{e}}{{m}_{\mu }}{m}_{e}{\alpha }^{4}{c}^{2}\\ =3.51218×{10}^{{}^{-6}}\text{eV}\end{array}$ (15-3)

4. 电子族磁矩反常现象

${\stackrel{^}{\mu }}_{i}=-{g}_{i}\frac{e}{2{m}_{i}}\stackrel{^}{S},\text{\hspace{0.17em}}\text{\hspace{0.17em}}i=e,\mu ,\tau$ (16)

4.1. 静电势

4.1.1. 中性粒子的受力问题

Figure 6. Comparison of electromagnetic and weak force

Table 1. Ratio of velocity of medium to light

4.1.2. 反向镜像

$\left\{\begin{array}{l}{\pi }_{±}\to {e}_{±}+\upsilon \\ {\pi }_{0}\to {e}_{±}+{e}_{\mp }\end{array}$ (17)

4.1.3. 电子族的衰变状态

$\phi \left(r\right)={\text{e}}^{-r/a},a=\alpha \frac{\hslash }{{m}_{e}c}$

Figure 7. Comparison of weak forces in two different types of decay

Figure 8. Reverse mirroring

$\phi \left(r\right)={\text{e}}^{-r/a},a=2\text{π}\alpha \frac{\hslash }{{m}_{e}c}$

$\phi \left(r\right)={\text{e}}^{-r/{a}_{i}},{a}_{i}=2\text{π}\alpha \frac{\hslash }{{m}_{i}c},\text{\hspace{0.17em}}i=e,\mu ,\tau$ (18)

${\stackrel{^}{E}}_{i}=-\frac{k{e}^{2}}{r}$ (19)

${\stackrel{¯}{E}}_{1}=-\left(\frac{\alpha }{2\text{π}}\right){m}_{i}{c}^{2}$ (20)

4.2. 动能

$\begin{array}{l}{{m}^{\prime }}_{e2}v=\frac{\alpha }{2\text{π}}{{m}^{\prime }}_{e}c\\ {\stackrel{¯}{E}}_{e2}=4×\frac{1}{2}\left({{m}^{\prime }}_{e2}\right){v}^{2}=2{\left(\frac{\alpha }{2\text{π}}\right)}^{2}{{m}^{\prime }}_{e}{c}^{2}\end{array}$ (21)

$\begin{array}{l}{m}_{\mu 2,\tau 2}v=\frac{\alpha }{2\text{π}}{m}_{\mu 2,\tau 2}c\\ {\stackrel{¯}{E}}_{\mu 2,\tau 2}=4×-\frac{1}{2}{m}_{\mu 2,\tau 2}{v}^{2}=-2{\left(\frac{\alpha }{2\text{π}}\right)}^{2}{m}_{\mu 2,\tau 2}{c}^{2}\end{array}$ (22)

4.3. 弱力势能

${\stackrel{¯}{E}}_{e3}=2×{\left(\frac{{m}_{e}}{{m}_{\mu }}\right)}^{3}{{m}^{\prime }}_{e}$ (23)

${\stackrel{¯}{E}}_{\mu 3}=4×{\left(\frac{{m}_{e}}{{m}_{\mu }}\right)}^{3}{m}_{\mu }$ (24)

${\stackrel{¯}{E}}_{\tau 3}=4×\left[{\left(\frac{{m}_{\mu }}{{m}_{\tau }}\right)}^{3}+{\left(\frac{{m}_{e}}{{m}_{\tau }}\right)}^{3}\right]{m}_{\tau }$ (25)

4.4. 电子族粒子之间的干涉

$\frac{\mathrm{Re}}{R\mu }=\frac{{m}_{e}^{2}{\left({m}_{x}^{2}-{m}_{e}^{2}\right)}^{2}}{{m}_{\mu }^{2}{\left({m}_{x}^{2}-{m}_{\mu }^{2}\right)}^{2}}$ (26)

$\mathrm{Re}\text{\hspace{0.17em}}+R\mu =1$ (27)

${m}_{x}$ 表示未知介子的质量，(26)、(27)包含了三个未知数，显然不能求解。我们可以带入缪子磁矩的实验值，反推求出 ${R}_{\mu }$ ；亦可通过电子磁矩反推求出 ${R}_{e}$

$\left\{\begin{array}{l}\mathrm{Re}=\frac{3}{8}\\ R\mu =\frac{5}{8}\end{array}$ (28)

 $x$ 介子的衰变反应有电子型 $\left(e,{\stackrel{¯}{\upsilon }}_{e}\right)$ 和缪子型 $\left(\mu ,{\stackrel{¯}{\upsilon }}_{\mu }\right)$ 两种类型(图9)。两种衰变反应衰变产物的自由组合共有$\left(e,{\stackrel{¯}{\upsilon }}_{\mu }\right)\cup \left(\mu ,{\stackrel{¯}{\upsilon }}_{e}\right)$$\left(e,\mu \right)\cup \left({\stackrel{¯}{\upsilon }}_{\mu },{\stackrel{¯}{\upsilon }}_{e}\right)$ 三种，其中第一种属于正常的衰变反应组合，第二、三种属于不同衰变反应之间衰变产物的自由组合，不同衰变反应之间衰变产物的自由组合占所有组合的份额为2/3。

${\stackrel{¯}{E}}_{e4}=\frac{8}{3}\cdot \frac{2}{3}{\alpha }^{4}\mathrm{Re}{{m}^{\prime }}_{e}{c}^{2}$ (29)

${\stackrel{¯}{E}}_{\mu 4}=\frac{8}{3}\cdot \frac{2}{3}{\alpha }^{4}\left(1-\mathrm{Re}\right){m}_{\mu }{c}^{2}$ (30)

${\stackrel{¯}{E}}_{\tau 4}=\frac{8}{3}{\alpha }^{4}{m}_{\tau }{c}^{2}$ (31)

4.5. 对冲效应

${{m}^{\prime }}_{e}=\left(1-\frac{{m}_{e}}{{m}_{p}}\right){m}_{e}$ (32)

Figure 9. Decay of x pion

${m}_{e0}=\left[1-\left(1-\frac{{m}_{e}}{{m}_{p}}\right)\left(\left(\frac{\alpha }{2\text{π}}\right)-2{\left(\frac{\alpha }{2\text{π}}\right)}^{2}+2{\left(\frac{{m}_{e}}{{m}_{\mu }}\right)}^{3}+\frac{16}{9}\mathrm{Re}{\alpha }^{4}\right)\right]{m}_{e}$ (33)

${m}_{\mu 0}=\left[1-\left(\left(\frac{\alpha }{2\text{π}}\right)+2{\left(\frac{\alpha }{2\text{π}}\right)}^{2}+4{\left(\frac{{m}_{e}}{{m}_{\mu }}\right)}^{3}+\frac{16}{9}\left(1-\mathrm{Re}\right){\alpha }^{4}\right)\right]{m}_{\mu }$ (34)

${m}_{\tau 0}=\left[1-\left(\left(\frac{\alpha }{2\text{π}}\right)+2{\left(\frac{\alpha }{2\text{π}}\right)}^{2}+4{\left(\frac{{m}_{\mu }}{{m}_{\tau }}\right)}^{3}+4{\left(\frac{{m}_{e}}{{m}_{\tau }}\right)}^{3}+\frac{8}{3}{\alpha }^{4}\right)\right]{m}_{\tau }$ (35)

${g}_{i}=-2×\frac{{m}_{i}}{{m}_{i0}},i=e,\mu ,\tau$ (36)

${g}_{e}=-2×{\left[1-\left(1-\frac{{m}_{e}}{{m}_{p}}\right)\left(\left(\frac{\alpha }{\text{2π}}\right)-2{\left(\frac{\alpha }{\text{2π}}\right)}^{2}+2{\left(\frac{{m}_{e}}{{m}_{\mu }}\right)}^{3}+\frac{16}{9}\mathrm{Re}{\alpha }^{4}\right)\right]}^{-}$ (37)

${g}_{\mu }=-2×{\left[1-\left(\left(\frac{\alpha }{\text{2π}}\right)+2{\left(\frac{\alpha }{\text{2π}}\right)}^{2}+4{\left(\frac{{m}_{e}}{{m}_{\mu }}\right)}^{3}+\frac{8}{3}\left(1-\mathrm{Re}\right){\alpha }^{4}\right)\right]}^{-}$ (38)

${g}_{\tau }=-2×{\left[1-\left(\left(\frac{\alpha }{\text{2π}}\right)+2{\left(\frac{\alpha }{\text{2π}}\right)}^{2}+4{\left(\frac{{m}_{\mu }}{{m}_{\tau }}\right)}^{3}+4{\left(\frac{{m}_{e}}{{m}_{\tau }}\right)}^{3}+\frac{16}{9}{\alpha }^{4}\right)\right]}^{-}$ (39)

5. 中微子的平面波方程组及转换

5.1. 中微子的平面波方程组

Table 2. g factor

Table 3. The figure of spread intermediary

1) 光子的平面波方程组：

$\begin{array}{l}\phi {\left(r,t\right)}_{\gamma }=\mathrm{exp}\left[-i\left(kr-\omega \stackrel{^}{t}\right)\right]\\ \stackrel{^}{E}=i\hslash \frac{\partial }{\partial t}\\ \stackrel{^}{P}=-i\hslash \frac{\partial }{\partial r}\\ r=x+y+z\end{array}$ (40)

2) 中微子的平面波方程组：

$\begin{array}{l}\phi {\left(r,t\right)}_{\upsilon }=\mathrm{exp}\left[-i\left(kr-\omega \stackrel{^}{t}\right)\right]\\ \stackrel{^}{E}=i\frac{\hslash }{2}\frac{\partial }{\partial t}\\ \stackrel{^}{P}=-i\frac{\hslash }{2}\frac{\partial }{\partial r}\\ r=x+y+z\end{array}$ (41)

$c=\frac{\omega }{k}$

5.2. 中微子转换

5.2.1. 太阳中微子失踪之谜

${\stackrel{¯}{\upsilon }}_{e}\to {\stackrel{¯}{\upsilon }}_{e}+\left\{\begin{array}{c}{\stackrel{¯}{\upsilon }}_{\mu }+{\upsilon }_{\mu }\\ {\stackrel{¯}{\upsilon }}_{\tau }+{\upsilon }_{\tau }\end{array}\to A+\left\{\begin{array}{c}{\stackrel{¯}{\upsilon }}_{\mu }\\ {\stackrel{¯}{\upsilon }}_{\tau }\end{array}$ (42)

A粒子与反缪(陶)子中微子的传播方向是重合的，当它们遇到原子核时，发生碰撞。在碰撞过程中，中微子可以跟原子核发生弱相互作用，这就是说，整个衰变及碰撞过程中，中微子的总数量保持不变，但粒子的总数量翻了一倍(图10)。

5.2.2. 偏振量定义

1) 偏振量e是一个与角度无关的常量，它在任意角度上的投影值均为1；

2) 在三种中微子偏振量之中，任意两两满足反对易关系。

${e}_{e},{e}_{\mu },{e}_{\tau }={\alpha }_{x},{\alpha }_{y},{\alpha }_{z}$ (43)

${P}_{e\to \mu ,\tau }=\frac{{e}_{\mu }^{2}+{e}_{\tau }^{2}}{{e}_{e}^{2}+{e}_{\mu }^{2}+{e}_{\tau }^{2}}=\frac{{\alpha }_{x}^{2}+{\alpha }_{y}^{2}}{{\alpha }_{x}^{2}+{\alpha }_{y}^{2}+{\alpha }_{2}^{2}}=\frac{2}{3}$ (44)

Figure 10. Neutrino’s conversion and capture process

Table 4. Decay and Combination of polarizations

5.2.3. 无中生有与中微子质量

${\pi }^{±}\to {\pi }^{0}+{e}^{±}+\upsilon$ 衰变反应如下图11所示。

π0介子的产生等价于无中生有一对正负π介子，相当于衰变逆反应。一句话总结π0介子产生的本质原因是由于正负π介子之间的磁矩相互作用(图12)。

${V}_{0\to {\pi }^{0}}=4{\left(\frac{{m}_{{\pi }^{0}}}{{m}_{{\pi }^{±}}}\right)}^{3}{m}_{{\pi }^{±}}{c}^{2}{\alpha }^{4}×{\left(1-0.5\mathrm{ln}\left(1+\alpha \right)\right)}^{-3}$ (45)

π0介子的产生概率等于正负π介子磁矩相互作用势/正负π介子质量：

$P=\frac{{V}_{0\to {\pi }^{0}}}{{m}^{\pi ±}{c}^{2}}=1.037×{10}^{-8}$ (46)

${\pi }^{±}\to {\pi }^{0}+{e}^{±}+\upsilon$ 衰变概率的实验值 [1] 为 $\left(1.025±0.034\right)×{10}^{-8}$ ，即两者相符。

6. 中微子的质量属性

Figure 11. ${\pi }^{±}\to {\pi }^{0}+{e}^{±}+\upsilon$

Figure 12. The magnetic moment interation between

The Research on Relationship between Neureinos and Weak Force[J]. 现代物理, 2017, 07(05): 197-212. http://dx.doi.org/10.12677/MP.2017.75023

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